Hi Mark, just wanted to share the solution to this problem as well as a method used.
I found quite a powerful tool buried at the end of this paper

"A Classification of Fully Residually Free Groups of Rank Three or Less",
Benjamin Fine, Anthony M Gaglione, Alexei Myasnikov, Gerhard Rosenberger, Dennis Spellman.

Here is the result that follows from the proofs/discussion of that paper and stated in the Section 8, Theorem 6:

Let $F_{u,v} = \langle u ,v \rangle $ be a free group of rank two and suppose that $a(u,v)$ and $b(u,v)$ are non-trivial and not proper powers in $F_{u,v}$. If $a(u,v)$ and $b(u,v)$ are conjugate in some overgroup $F_{u,v} < F$ then either $a(u,v)$ and $b(u,v)$ are already conjugate in $F_{u,v}$ or the element $ta(u,v)t^{-1}b(u,v)^{-1}$ is a primitive in a free group of rank three $F_{3} = \langle t,u,v \rangle $.

Let me now demonstrate that the equation $[x,u]=[u,v]$ has no solutions in $F$.
Suppose the above equation has a solution $x=\omega$. We have then $\omega^{-1}u^{-1}\omega =[u,v]u^{-1}$, so that elements $u^{-1}$ and $[u,v]u^{-1}$ are conjugated in $F$. First, notice that $u^{-1}$ and $[u,v]u^{-1}$ are neither proper powers nor conjugated in $F_2= \langle u,v \rangle $.
Hence, by the above result, element $e=t^{-1}u^{-1}tu[v,u]$ must be primitive in $F_3 = \langle t,u,v \rangle$. However element $e$ belongs to the derived subgroup $F_3 '$ and as such can not be primitive.